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3 years ago

PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer: x=10

Step-by-step explanation:

x/20=14/28

28x=14 times 28

x=10

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After releasing of radioactive material into.The atnosphere from.A nuclear.Powe plant in a country in 1988, the hay in that coun

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Answer:

the question is incomplete, so I looked for a similar one:

<em>After the release of radioactive material into the atmosphere from a nuclear power plant, the hay was contaminated by iodine 131 ( half-life, 8 days). If it is all right to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? </em>

iodine's half life (we are given x, we need to find b):

0.5A₀ = A₀eᵇˣ

x = 8 days

we eliminate A₀ from both sides

0.5 = eᵇ⁸

ln 0.5 = ln eᵇ⁸

-0.69315 = b8

b = -0.69315 / 8 = -0.08664

since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):

0.1A₀ = A₀eᵇˣ

0.1 = eᵇˣ

ln 0.1 = bx

where b = -0.08664

x = ln 0.1 / -0.08664 = -2.302585 / -0.08664 = 26.58 days

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3 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

Oscar invests $20,000 in three investments earning 6% ,8% and 10%. He invests $9000 more in the 10% investment than in the 6% in

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This problem is mainly concerned with setting up an equation.

let's add the interest rates times their respective investment amounts:

$a = (9000 + x)0.10$

$b = x \times 0.06$

$c=(20000 - ((9000 + x) + x)) 0.08$

simplify:

$c = (11000 - 2x) \times 0.08$

sum of interest:

$1780 = a + b + c$

1780 = (9000+x) 0.10 + x 0.06 + (11000-2x)0.08

$1780 = 1780 + 0 \times x$

$0=0$

solution:

He does not have any invested in the 6%, while he has 9000 invested at 10% and 11,000 invested at 8%

to check:

$9000 \times 0.10 + 11000 \times 0.08$

$= 900 + 880 = 1780$

So this ended up being a trick problem because he does not actually have any invested in the 6% option.

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3 years ago